If $x_1,x_2,...,x_{2017} \in\Bbb{R}^+$ and [MATH]\dfrac{1}{1+x_1}+\dfrac{1}{1+x_2}+...+\dfrac{1}{1+x_{2017}} = 1[/MATH], find the minimal possible value of [MATH]\prod_{i=1}^{2017}x_i[/MATH].

**Solution:**
By cyclic symmetry, we know the critical value is at the point:

[MATH]\left(x_1,\cdots,x_{2017}\right)=(2016,\cdots,2016)[/MATH]

And the objection function at that point is:

[MATH]f(2016,\cdots,2016)=2016^{2017}[/MATH]

Now, looking at another point on the constraint:

[MATH]\left(4032,4032,\cdots,4032,\frac{2016}{2017}\right)[/MATH]

We find the objective function at that point is:

[MATH]f\left(4032,4032,\cdots,4032,\frac{2016}{2017}\right)=\frac{2^{2016}2016^{2017}}{2017}>2016^{2017}[/MATH]

And so we conclude:

[MATH]f_{\min}=2016^{2017}[/MATH]

[MATH]\left(x_1,\cdots,x_{2017}\right)=(2016,\cdots,2016)[/MATH]

And the objection function at that point is:

[MATH]f(2016,\cdots,2016)=2016^{2017}[/MATH]

Now, looking at another point on the constraint:

[MATH]\left(4032,4032,\cdots,4032,\frac{2016}{2017}\right)[/MATH]

We find the objective function at that point is:

[MATH]f\left(4032,4032,\cdots,4032,\frac{2016}{2017}\right)=\frac{2^{2016}2016^{2017}}{2017}>2016^{2017}[/MATH]

And so we conclude:

[MATH]f_{\min}=2016^{2017}[/MATH]